Question

A particle moves on xy plane. Its position vector at any time t is $\overrightarrow{r}=\left\{\right(2t)\hat{i}+(2t^2\left)\hat{j}\right\}m$. The rate of change of $\theta$ at time t=2 second. (Where $\theta$ is the angle which its velocity vector makes with positive x-axis) is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\overrightarrow{r}=2t\hat{i}+2t^2\hat{j}$
Differentiating w.r.t t to get velocity,
$\overrightarrow{v}=2\hat{i}+4t\hat{j}$
Angle that velocity makes with x-axis is slope =$tan\theta =\frac{4t}{2}=2t.....\left(1\right)$

Now, we have to find $\frac{d\theta }{dt}$ at t=2s, differentiating equation (1) we get
$se{c}^2\theta \frac{d\theta }{dt}=2$
$\frac{d\theta }{dt}=\frac{2}{se{c}^2\theta }$
$\frac{d\theta }{dt}=\frac{2}{1+ta{n}^2\theta }=\frac{2}{1+4t^2}$ We know $se{c}^2\theta =1+ta{n}^2\theta$
at t=2
$\frac{d\theta }{dt}=\frac{2}{1+16}=\frac{2}{17}rad/sec$